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\newtheorem{theorem}{Theorem}[section]%定理
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\title{\heiti\zihao{2} 习题15.6}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{求 $f(x, y)=x^{3}+3 x y^{2}-15 x-12 y$ 的极值.}
\textbf{解}\quad
分别对$x,y$求偏导得:
$$
\begin{aligned}
    \dfrac{\partial f}{\partial x}&=3x^2+3y^2-15=0\\
    \dfrac{\partial f}{\partial y}&=6xy-12=0
\end{aligned}
$$
即$\left\{\begin{array}{l}
    x^2+y^2=5\\
    xy=2
\end{array}\right.$.解得$(x,y)=\left\{\begin{array}{l}
    (1,2)\\
    (-1,-2)\\
    (2,1)\\
    (-2,-1)
\end{array}\right.$.分别在四个点求海色矩阵得
$$
\begin{aligned}
    (1,2)&:\left[\begin{array}{cc}
        6x & 6y\\
        6y & 6x
    \end{array}\right]=\left[\begin{array}{cc}
        6 & 12\\
        12 & 6
    \end{array}\right]\\
    (-1,-2)&:\left[\begin{array}{cc}
        6x & 6y\\
        6y & 6x
    \end{array}\right]=\left[\begin{array}{cc}
        -6 & -12\\
        -12 & -6
    \end{array}\right]\\
    (2,1)&:\left[\begin{array}{cc}
        6x & 6y\\
        6y & 6x
    \end{array}\right]=\left[\begin{array}{cc}
        12 & 6\\
        6 & 12
    \end{array}\right]\\
    (1,2)&:\left[\begin{array}{cc}
        6x & 6y\\
        6y & 6x
    \end{array}\right]=\left[\begin{array}{cc}
        -12 & -6\\
        -6 & -12
    \end{array}\right]\\
\end{aligned}
$$
显然$(1,2),(-1,-2)$处的海色矩阵都是不定矩阵,所以在这两点不是极值.在$(2,1)$处的海色矩阵是正定矩阵,所以该点处为极小值,为$-28$.在$(2,1)$处的海色矩阵是负定矩阵,所以该点处为极大值,为$28$.
\section{求函数 $f(x, y)=x^{3}-y^{3}+3 x^{2}+3 y^{2}-9 x$ 的极值.}
\textbf{解}\quad
分别对$x,y$求偏导得:
$$
\begin{aligned}
    \dfrac{\partial f}{\partial x}&=3x^2+6x-9=0\\
    \dfrac{\partial f}{\partial y}&=-3y^2+6y=0
\end{aligned}
$$
解得$(x,y)=\left\{\begin{array}{l}
    (1,0)\\
    (1,2)\\
    (-3,0)\\
    (-3,2)
\end{array}\right.$.分别在四个点求海色矩阵得
$$
\begin{aligned}
    (1,0)&:\left[\begin{array}{cc}
        6x+6 & 0\\
        0 & -6y+6
    \end{array}\right]=\left[\begin{array}{cc}
        12 & 0\\
        0 & 6
    \end{array}\right]\\
    (1,2)&:\left[\begin{array}{cc}
        6x+6 & 0\\
        0 & -6y+6
    \end{array}\right]=\left[\begin{array}{cc}
        12 & 0\\
        0 & -6
    \end{array}\right]\\
    (-3,0)&:\left[\begin{array}{cc}
        6x+6 & 0\\
        0 & -6y+6
    \end{array}\right]=\left[\begin{array}{cc}
        -12 & 0\\
        0 & 6
    \end{array}\right]\\
    (-3,2)&:\left[\begin{array}{cc}
        6x+6 & 0\\
        0 & -6y+6
    \end{array}\right]=\left[\begin{array}{cc}
        -12 & 0\\
        0 & -6
    \end{array}\right]\\
\end{aligned}
$$
显然$(1,2),(-3,0)$处的海色矩阵都是不定矩阵,所以在这两点不是极值.在$(1,0)$处的海色矩阵是正定矩阵,所以该点处为极小值,为$-5$.在$(-3,2)$处的海色矩阵是负定矩阵,所以该点处为极大值,为$31$.

\section{求函数 $z=x^{2}+5 y^{2}-6 x+10 y+6$ 的极值点.}
\textbf{解}\quad
分别对$x,y$求偏导得:
$$
\begin{aligned}
    \dfrac{\partial f}{\partial x}&=2x-6=0\\
    \dfrac{\partial f}{\partial y}&=10y+10=0
\end{aligned}
$$
解得$(x,y)=(3,-1)$.求海色矩阵得
$$
\begin{aligned}
    \left[\begin{array}{cc}
        2 & 0\\
        0 & 10
    \end{array}\right]
\end{aligned}
$$
显然$(3,-1)$处的海色矩阵是负定矩阵,所以该点处为极大值,为$-8$.

\section{考察 $z=-\sqrt{x^{2}+y^{2}}$ 是否有极值.}
\textbf{解}\quad
$z\leqslant 0$.在$(0,0)$附近,都有$-\sqrt{x^2+y^2}<0$,所以$(0,0)$为极大值.而对$x,y$分别求偏导,并令偏导数为$0$得$(x,y)=(0,0)$.所以只有极大值,没有极小值.极大值在$(0,0)$处取得$0$.

\section{在椭球面 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}=1$ 的第一卦限求一点,使该点的切平面与三坐标面围成的四面体的体积最小.}
\textbf{解}\quad
设第一象限点为$(x_0,y_0,z_0)$,则在该点的法向量$\vec{n}=\left(\dfrac{2x_0}{a^2},\dfrac{2y_0}{b^2},\dfrac{2z_0}{c^2}\right)$.所以该点处的切平面为$\dfrac{2x_0}{a^2}(x-x_0)+\dfrac{2y_0}{b^2}(y-y_0)+\dfrac{2z_0}{c^2}(z-z_0)=0$.\par 
与$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}=1$联立,求出该平面与三个坐标轴的截距为$A=\dfrac{a^2}{x_0},B=\dfrac{b^2}{y_0},C=\dfrac{c^2}{z_0}$.而体积为$\dfrac{1}{6}ABC=\dfrac{a^2b^2c^2}{6x_0y_0z_0}$.问题化为求$x_0y_0z_0$的最大值.\par 
令$f(x,y,z)=xyz-\lambda\left(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}-1\right)$.则由Lagrange乘数法得:
$$
\begin{aligned}
    &\dfrac{\partial f}{\partial x}=yz-\dfrac{2\lambda x}{a^2}=0\\
    &\dfrac{\partial f}{\partial y}=xz-\dfrac{2\lambda y}{b^2}=0\\
    &\dfrac{\partial f}{\partial z}=xy-\dfrac{2\lambda z}{c^2}=0\\
    &\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}=1
\end{aligned}
$$
可知$x^2y^2z^2=\dfrac{8\lambda^3xyz}{a^2b^2c^2}$.所以$xyz=\dfrac{8\lambda^2}{a^2b^2c^2}$.从而解得$\lambda = \dfrac{\sqrt{3}abc}{6},x=\dfrac{\sqrt{3}a}{3},y=\dfrac{\sqrt{3}b}{3},z=\dfrac{\sqrt{3}c}{3}$.\par 
该点处得海色矩阵:
$$
\left[\begin{array}{ccc}
    -\dfrac{\sqrt{3}bc}{3a} & 0 & 0\\
     0& -\dfrac{\sqrt{3}ca}{3b} & 0\\
     0& 0 & -\dfrac{\sqrt{3}ab}{3c}\\
\end{array}\right]
$$
是负定矩阵,所以该点取$xyz$最大值,所以取$V$的最小值.\par
综上,该点为$\left(\dfrac{\sqrt{3}a}{3},\dfrac{\sqrt{3}b}{3},\dfrac{\sqrt{3}c}{3}\right)$.


\section{求函数 $z=x^{2}+y^{2}$ 在圆 $(x-\sqrt{2})^{2}+(y-\sqrt{2})^{2} \leqslant 9$ 上的最大值与最小值.}
\textbf{解}$1^{\circ}$\quad
几何做法,只需求在以圆心为$(\sqrt{2},\sqrt{2})$,半径小于等于$3$的圆系中与原点距离最大与最小的值的平方.显然半径为$2$的时候能够取到最小值$0$,半径为$3$的时候能取到最大值$25$.\par 

\textbf{解}$2^{\circ}$\quad
设$f(x,y)=x^2+y^2-\lambda\left[(x-\sqrt{2})^2+(y-\sqrt{2})^2-9\right]$.由Lagrange乘数法可得:
$$
\begin{aligned}
    \dfrac{\partial f}{\partial x}&=2x-2\lambda(x-\sqrt{2})=0\\
    \dfrac{\partial f}{\partial y}&=2y-2\lambda(y-\sqrt{2})=0\\
    (x-\sqrt{2})^2+(y-\sqrt{2})^2&\leqslant 9
\end{aligned}
$$
解得$x=y=\dfrac{\lambda\sqrt{2}}{\lambda -1}$.将其带入$x^2+y^2$,得$\lambda = 0$时其取到最小值,且海色矩阵为$\left[\begin{array}{cc}
    2-2\lambda & 0\\
    0 & 2-2\lambda
\end{array}\right]$.是正定矩阵.所以极小值为$0$.\par 
将其带入$(x-\sqrt{2})^2+(y-\sqrt{2})^2\leqslant 9$可得$|\lambda -1|\geqslant \dfrac{2}{3}$.再将此条件带入$z=x^2+y^2$可得$z<\leqslant \dfrac{4(\dfrac{5}{3})^2}{(\dfrac{2}{3})^2}=25$.当$x=y=\dfrac{5\sqrt{2}}{2}$时成立,且海色矩阵为$\left[\begin{array}{cc}
    -\dfrac{4}{3} & 0\\
    0 & -\dfrac{4}{3}
\end{array}\right]$,是负定矩阵,所以为极大值.考虑边界情况,并综上,易得最小值为$0$,最大值为$25$.

\section{求 $g(x, y)=1-x^{2/3}-y^{4/5}$ 的极值.}
\textbf{解}\quad
分别对$x,y$求偏导得:
$$
\begin{aligned}
    \dfrac{\partial g}{\partial x}&=-\dfrac{2}{3}x^{-1/3}=0\\
    \dfrac{\partial g}{\partial y}&=-\dfrac{4}{5}y^{-1/5}=0
\end{aligned}
$$
$(0,0)$处海色矩阵$\left[\begin{array}{cc}
    \dfrac{2}{9} & 0\\
    0 & \dfrac{4}{25}
\end{array}\right]$显然为正定矩阵,所以该点为极大值点.
\end{document}